最小距离求解

Created on: January 6, 2025

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Question

已知椭圆C:\frac {x^{2}}{8}+y^{2}=1，定点P(5,6)，求定点P到椭圆C最近距离。要求：“
按照高中数学作答规范

首先分析题目类型,明确解题要求
列出涉及的关键知识点
按照完整的解题步骤进行详细解答:
1. 给出解题思路和方法选择的说明
2. 展示规范的演算过程
3. 注意数学符号使用的准确性
4. 对重要步骤进行必要的解释”

Thought

Mapping out the solution

I’m working through finding the shortest distance from a point to an ellipse using Lagrange multipliers to deal with the constraint.

Solving the equations

I'm working through two equations to find λ. Then, substituting y in terms of x into the ellipse equation to connect them, leading to a combined equation.

Adjusting my approach

I’m simplifying the equation by clearing denominators and expressing terms with a substitution. This results in a polynomial equation in terms of x.

Unveiling the quartic

I expanded the polynomial, simplified terms, and combined like terms to form a quartic equation. Now, I’m checking if x=0 is a solution.

Testing potential solutions

I’m mapping out x=5 as a solution to the quartic equation. Then, I opt for a geometric method, parameterizing the ellipse and minimizing the distance from a point on the ellipse to (5,6).

Crafting a strategy

I’m expanding f(t) and using trigonometric identities. Recognizing cos^2 t + sin^2 t = 1 simplifies the expression to 7 cos^2 t + 1.

Taking derivatives and factoring

I’m differentiating f(t) with respect to t, aiming to find critical points. Factoring out sin t or cos t could simplify the equation, potentially revealing solutions.

Solving sine and cosine

I’m factoring and examining critical points of $f'(t)$ . If $\cos t = 0$ , the nearby ellipse points might not yield the minimal distance to $P(5,6)$ .

Working through trigonometric relationships

OK, I’m defining $\tan t = k$ to express $\sin t$ and $\cos t$ in terms of $k$ , which could lead to an optimal solution.

Revisiting trigonometric equations

I’m re-evaluating the trigonometric substitution, aiming to balance terms and simplify. Ensuring both sine and cosine components are precisely managed is key to finding an optimal solution.

Simplifying expressions

OK, let me see. I’m mapping out trigonometric relationships using a single variable $k = \tan t$ . This should streamline the equation and make the problem more manageable.

Here’s what I’m working on

I’m trying out different trigonometric manipulations to express $\cos t$ and $\sin t$ in simpler terms, aiming for a straightforward, solvable trigonometric equation.

Piecing together equations

Substituting $\cos t$ and $\sin t$ into their relationship, I’m arriving at a quartic equation in terms of $s$ , aiming to identify potential solutions.

Trying another way

I’m digging into different methods to solve the quartic equation, including numeric and geometric approaches. It’s interesting to see how these strategies might simplify the problem.

Taking a closer look

I'm focusing on $f(t) = (2 \sqrt{2} \cos t - 5)^2 + (\sin t - 6)^2$ , aiming to find critical points where $f'(t) = 0$ .

Taking a closer look

I’m aiming to differentiate $f(t)$ with respect to $t$ , focusing carefully on each term. This approach will help pinpoint critical points where $f'(t) = 0$ .

Piecing together derivatives

I’m working through the derivative of $f(t)$ , careful with each term. This allows me to pinpoint critical points effectively.

Piecing together terms

I’m differentiating $f2(t)$ with respect to $t$ , skipping intermediate steps. This helps identify critical points where $f1'(t) = 0$ and $f2'(t) = 0$ .

Grouping and simplifying

I’m grouping and factoring terms, setting the derivative to zero. This reveals critical points by equating the result to zero, using trigonometric identities and substitutions.

Assuming $\cos t$ as $\frac{1}{\sqrt{1 + \tan^2 t}}$ : This definition helps simplify the equation and align with trigonometric identities.

Reworking the equation

I’m expressing $k$ and $M$ to simplify the equation, aiming to balance terms and align them with trigonometric identities. This makes finding solutions more straightforward.

Piecing together

I’m expanding and simplifying terms to form a quartic equation in $k$ . This process involves grouping and balancing terms with $\sqrt{2}$ and other elements.

Testing rational solutions

OK, let me see: I’m assessing $k = 1$ and $k = 3$ as potential rational solutions for the quartic equation. However, these values don’t seem to satisfy the equation.

Defining the problem

I'm working through the challenge of minimizing the distance from $P(5,6)$ to the ellipse, focusing on an approximate numerical method to find the nearest point on the ellipse boundary.

Mapping out distance

I’m testing different parameter values for the ellipse to find the nearest point to $P(5,6)$ . Values for $t$ like $t = 0$ , $t = \pi/2$ , and $t = \pi/4$ are considered.

Pinpointing the optimal angle

Testing $t = 0.9$ radians, I’m calculating the coordinates and distance. This refinement should more accurately represent the closest point on the ellipse to $P(5,6)$ .

Testing angles

I'm refining values for $t$ to hone in on the closest point on the ellipse to $P(5,6)$ . Adjustments in increments of $t$ are revealing clearer results.

Mapping out angles

Mapping multiple angles, it appears the minimal distance hovers around 6.04-6.06. Continuously refining the angle should accurately determine the closest point on the ellipse to $P(5,6)$ .

Determining key values

I’m concluding that the minimal distance from $P(5,6)$ to the ellipse is likely around 6.04. Considering the problem requirements, a numeric or simple exact solution seems plausible.

Calculating critical points

I’m mapping out the derivative approach to pinpoint critical points, refining my guess for $t$ to narrow down the minimal distance.

Nailing down critical points

Piecing together the derivative for $t$ around 0.73 to pinpoint minimal distance. The positive derivative suggests decreasing $t$ could lead to a smaller function value.

Tracking progression

I’m honing in on a precise interval for $t$ and evaluating the sum of terms. This ensures closer adherence to the guideline formatting.

Estimating near the root

I'm narrowing down the interval for $t$ to pinpoint the root. The sign shifts between $t=0.622$ and $t=0.623$ , suggesting the root is between these values.

Nailing the root

I’m honing in on the root for $t$ by testing values between $t=0.622$ and $t=0.623$ . This helps confirm the critical point more accurately.

Estimating the distance

Testing various $t$ values, I find that the minimal distance from $P(5,6)$ to the ellipse is roughly 6.057, when using $t$ around 0.63.

Evaluating angle proximity

I’m thinking through the closeness of angles to pinpoint the minimal distance from $P(5,6)$ to the ellipse, concluding that $t$ around 0.63 yields a distance of about 6.05.

Thought about 椭圆与距离计算，持续 1m 38s

Answer

下面给出一个按照高中数学“规范化解题”要求所写的详细示范。题目如下：

已知椭圆 $C: \frac{x^2}{8} + y^2 = 1$ ，定点 $P(5,6)$ 。求定点 $P$ 到椭圆 $C$ 的最近距离。

一、题目类型与解题要求

题目类型
这是一个“最值”类的几何优化问题，需要在给定曲线（椭圆）上，寻找一点使得到定点 $P$ 的距离最小。
解题要求
- 根据椭圆方程 $\frac{x^2}{8} + y^2 = 1$ ，在其上寻找使 $|PP'|$ 最小的点 $P'$ 。
- 最终需要给出“最近距离”的数值结果（如若能得到精确式则最好，若式子过于复杂，也可给出数值近似）。
- 解题过程要有条理，并注意高中阶段的书写规范、数学符号的准确使用以及重要步骤的解释。

二、关键知识点

椭圆的标准方程与参数方程
给定椭圆 $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ 。本题中 $a^2 = 8$ ， $b^2 = 1$ ，所以 $a = 2\sqrt{2},\, b = 1$ 。
常用的参数方程可写为
$x = a \cos t = 2\sqrt{2} \cos t,\quad y = b \sin t = \sin t.$
最值问题常用方法
- 方法 1：拉格朗日乘数法
  适用于带约束的优化问题： $\min\; f(x,y) \quad\text{约束：}\; g(x,y)=0.$ 求解 $\nabla f = \lambda \nabla g$ 及约束方程。
- 方法 2：参数法
  将椭圆上的点用单一参数 $t$ 表示，再把目标函数（距离或距离的平方）转化为只含 $t$ 的函数，求导数为零即可。
求距离最小等价于求距离的平方最小
在做最值运算时，为了简化运算，通常将
$\min \sqrt{(x-5)^2 + (y-6)^2}$
等价替换为
$\min \bigl[(x-5)^2 + (y-6)^2\bigr].$

三、解题思路与方法选择

由于使用拉格朗日乘数法会得到较为繁琐的方程组(往往会化为四次方程)，且最后往往只能数值求解。
在高中常见的做法是：用椭圆的参数方程 $x = 2\sqrt{2}\cos t,\; y = \sin t$ ，再用“导数=0”来求极值，会更直接一些。

下面示范参数法的详细步骤。

四、详细解题过程

1. 设椭圆上任意一点的参数表示

令

x = 2\sqrt{2}\cos t,\quad y = \sin t,

其中 $t$ 为实数参数。这样任意 $(x,y)$ 都满足 $\frac{x^2}{8} + y^2 = 1$ 。

2. 写出目标函数（距离的平方）

我们要最小化点 $P'(x,y)$ 与 $P(5,6)$ 的距离。令

D^2 = (x - 5)^2 + (y - 6)^2.

带入参数方程：

x - 5 = 2\sqrt{2}\cos t - 5,\quad y - 6 = \sin t - 6.

因此

D^2(t) = \bigl(2\sqrt{2}\cos t - 5\bigr)^2 \;+\; \bigl(\sin t - 6\bigr)^2.

3. 将 $D^2(t)$ 展开并求导

展开
$D^2(t) = \bigl(2\sqrt{2}\cos t - 5\bigr)^2 + (\sin t - 6)^2.$
为方便后续求导，这里并不一定要完全展开，而是直接对其做导数亦可。
对 $D^2(t)$ 求导
记
$f(t) = D^2(t).$
则
$f'(t) = \frac{d}{dt}\Bigl[(2\sqrt{2}\cos t - 5)^2\Bigr] + \frac{d}{dt}\Bigl[(\sin t - 6)^2\Bigr].$
- 第一项 $\frac{d}{dt}\Bigl[(2\sqrt{2}\cos t - 5)^2\Bigr] = 2\,(2\sqrt{2}\cos t - 5)\,\frac{d}{dt}(2\sqrt{2}\cos t - 5) = 2\,(2\sqrt{2}\cos t - 5)\,(2\sqrt{2})(-\sin t).$ 整理后可写成 $-4\sqrt{2}\,\bigl(2\sqrt{2}\cos t - 5\bigr)\sin t.$
- 第二项 $\frac{d}{dt}[(\sin t - 6)^2] = 2(\sin t - 6)\cos t.$
因此，
$f'(t) = -4\sqrt{2}\,\bigl(2\sqrt{2}\cos t - 5\bigr)\sin t \;+\; 2(\sin t - 6)\cos t.$
令导数 $f'(t) = 0$ ，求极值
要得到距离最小，就需要令
$f'(t) = 0.$
这会得到一个关于 $\sin t$ 和 $\cos t$ 的方程。从实际计算来看，它最终会化为一个四次方程，解析解极其繁琐，一般需要数值方法逼近。

在高中竞赛或高考延伸中，若该方程没有简单根，往往最后给出一个近似数值解即可。

4. 数值逼近并求得最小距离

通过对方程

f'(t) = 0

进行数值求解(或在没有计算器时做一些“试值+分析”，或者用更高阶工具/算法)，我们大致可以找到使 $D^2$ 最小的 $t \approx 0.62 \sim 0.63$ 之间。

以 $t\approx 0.63$ 为例： $x \approx 2\sqrt{2}\cos(0.63), \quad y \approx \sin(0.63).$ 数值上约为 $x \approx 2.28,\quad y \approx 0.59.$ 这时 $D^2 \approx (2.28 - 5)^2 + (0.59 - 6)^2 \approx 7.37 + 29.26 \approx 36.63,$ 因而 $D \approx \sqrt{36.63} \approx 6.05.$

综合更精细的数值搜索，得到的最小距离大约在

\boxed{6.05 \text{（约）}}

左右。

五、结果与结论

最小距离数值
由数值计算可知，定点 $P(5,6)$ 到椭圆 $\frac{x^2}{8} + y^2 = 1$ 的最小距离约为
$\boxed{6.05}.$
方法评价
- 使用参数法在高中较为常见，可以直接对距离或距离的平方进行“一元函数求导”来求最小值。
- 若追求解析解，将面临四次方程，代数处理相对复杂，最终也多是数值结果。
几点说明
- 如果考试或竞赛允许使用计算器，直接通过数值方法或结合必要的试值、逼近方法来求得最优 $t$ ，再得到最小距离。
- 在并不要求严格解析表达式的场合，给出准确的“数值解”是足够的。
- 若某些场合需精确解，则会得到一个比较繁琐的根式形式（四次方程解），不在常规高中解法之列。

最终答案

\text{定点 }P(5,6)\text{ 到椭圆 } \frac{x^2}{8} + y^2 = 1 \text{ 的最近距离约为 } \boxed{6.05}.